Question

A 100​-watt ​[W] motor ​(70​% ​efficient) is used to raise a 100​-kilogram ​[kg] load 5 meters​ [m] into the air. How​ long, in units of seconds​ [s], will it take the motor to accomplish this​ task?

Answer 1

Answer:

t= 71.42 s

Explanation:

Given that

Power ,P = 100 Watt

Efficiency of the motor ,η = 70 %

mass , m = 100 kg

h = 5 m

Let's take time taken = t

The work done to move 100 kg by 5 m

W= m g h

W= 100 x 10 x 5           ( take g= 10 m/s²)

W = 5000 J

We know that  1 Watt= 1 J/s

Therefore

P x t= W

Now by putting the values in the above equation

100 x t x 0.7= 5000

$t=\dfrac{5000}{70}\ s$

t= 71.42 s

Therefore the time taken by motor will be 71.42 s.